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A369110 a(n) is the number of distinct elements appearing in the sequence formed by recursively applying A063655 when starting from n.

%I A369110 #29 Apr 25 2024 13:24:11
%S A369110 4,3,2,1,2,2,4,3,3,5,6,5,5,4,4,4,5,4,5,4,6,6,7,6,6,5,6,7,8,7,7,6,5,6,
%T A369110 6,6,8,7,5,6,7,6,6,5,5,7,6,5,5,5,5,6,6,5,5,5,7,8,6,5,7,6,5,5,5,6,8,7,
%U A369110 6,6,7,6,7,6,5,8,5,6,6,5,5,7,7,6,7,6,7,6,7,6,5,7,7,6,7,5,8,7,5
%N A369110 a(n) is the number of distinct elements appearing in the sequence formed by recursively applying A063655 when starting from n.
%C A369110 A063655(n) gives the smallest semiperimeter of an integral rectangle with area n, which is the same thing as the minimum sum of two positive integers whose product is n. In this sequence, A063655 is applied recursively until a cycle is found. Then the number of distinct elements appearing in this process is given as a(n). Note that it's conjectured that a cycle will be found at some point.
%C A369110 Conjecture: The cycle part of each sequence generated by the recursion is one of (4), (5, 6), or (6, 5). Confirmed through 1 millionth term.
%C A369110 The conjecture is true. Proof: The conjecture holds for n <= 6. Suppose n >= 7 and the conjecture holds for lower values of n. If n is composite, then A063655(n) <= n/2+2. If n is prime, then A063655(n) = n+1 is even and A063655(A063655(n)) <= (n+1)/2+2. In both cases, n reaches a lower number and the conjecture holds for n. - _Jason Yuen_, Mar 30 2024
%H A369110 Jason Yuen, <a href="/A369110/b369110.txt">Table of n, a(n) for n = 1..10000</a>
%e A369110 n = 1 can be factored as 1*1 with minimum sum 2 (similarly, A063655(1) = 2). Then 2 = 1*2, so minimum sum is 3 = A063655(2). 3 = 1*3 which means the next number in the recursion is 4 = A063655(3). 4 = 2*2 which gives the same number 4 = A063655(4), hence this recursion will create a cycle at this point. Starting from n = 1 (including 1), we generated these numbers: (1, 2, 3, 4, 4, 4, ...). Therefore, a(1) = 4. a(2), a(3), and a(4) are trivially deduced from this example.
%o A369110 (Python)
%o A369110 from sympy import divisors
%o A369110 def A369110(n):
%o A369110     c = {n}
%o A369110     while n<4 or n>5:
%o A369110         c.add(n:=(d:=divisors(n))[((l:=len(d))-1)>>1]+d[l>>1])
%o A369110     if n==5:
%o A369110         c.add(6)
%o A369110     return len(c) # _Chai Wah Wu_, Apr 25 2024
%Y A369110 Cf. A063655, A056737.
%K A369110 nonn,easy
%O A369110 1,1
%A A369110 _Adnan Baysal_, Jan 13 2024